See L1 value properly and solve
please. Many just submit ready made solution without
reading the question.
L1 value is bm a cn .. CHECK
IT..
Solve B first if you know. Else not then C. Thank
you.
Problem B. Use the minimum state lemma to prove that the language L2={bm a c" |n>m>0} is not regular. First, define a set D that contains an infinite number of strings that are pairwise distinguishable with respect to L. Then show that there is a distinguisher for any arbitrary pair of strings from D. Write precisely and unambiguously. The
an answer sheet.
For problems C and D, use the pumping lemma to prove that Li and L2 are not regular by giving an appropriate answer for each blank in the given incomplete proofs. Use the example proofs given in the class as a reference and follow the guidelines stated below. For the blank (1), write a string precisely using the alphabet {a, b} and the constant p, e.g., ap+10 b2p+999. Note this string must be of length p or longer.
example, ab2n for some integer n 99. Make sure that any variable or constant used is defined clearly, such as for some
conditions stated in the incomplete proof are satisfied. For the blank (3), write an integer constant as the exponent in y2, such as 0, 5 or 9999999. For the blank (4), write the pumped sentence of xyz as a string precisely using the alphabet {a,b,c} and possibly integer variables and constants, e.g., ap-5k p2p+8 . Make sure that all variables and constants used are defined. For the blank (5), state a reason precisely, for example, p -5k < p for any k>0, or, 10k+1 cannot be even for any integer k. Make sure that all variables and constants used are clearly defined.
Problem C. Prove that L1 = {bi ci am|i, j > 0 and m=i+j } is not regular. Proof. For contradiction, assume that Li is regular. By the pumping lemma there exists a constant p > 0 such that all sentences of Li of a length > p can be pumped. Consider the sentence s= (1)_, whose length is > p. No matter how we divide s into three parts x, y, z such that s=xyz, if (yl>0, and (xyl p, then the part y must be (2). Hence, xy-(3) z =4_, which is not in L because 5_. That is, the language L1 contradicts the pumping lemma and, hence, is not regular. Type your answer for each blank on an answer sheet.
